Difference between revisions of "HW1.3 Kofo Adafin - Inverse of Complex Numbers ECE301Fall2008mboutin" - Rhea

Revision as of 04:27, 2 September 2008

${(a+bi)^{-1}}={1\over(a + bi)} = {a\over a^2+b^2}+ \left( {-b\over a^2+b^2}\right)i$

Revision as of 04:23, 2 September 2008 (view source) Kadafin (Talk) (→‎Finding the Inverse of Complex Numbers) ← Older edit		Revision as of 04:27, 2 September 2008 (view source) Kadafin (Talk) (→‎Finding the Inverse of Complex Numbers) Newer edit →
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	== Finding the Inverse of Complex Numbers ==		== Finding the Inverse of Complex Numbers ==

−	<math>1/(a + bi) = {a\over a^2+b^2}+ \left( {-b\over a^2+b^2}\right)i</math>	+	<math>{(a+bi)^{-1}}={1\over(a + bi)} = {a\over a^2+b^2}+ \left( {-b\over a^2+b^2}\right)i</math>